Let $X$ be the vector space of all Lebesgue-measurable functions $f:\left[a,b\right]\rightarrowℝ$ such that $\int^{b}_{a}\left|f\left(x\right)\right|^{2}dx<\infty$ (Lebesgue integral). Then we can define an equivalence relation on $X$ as follows: $f \cong g$ if $f(x)=g(x)$ almost everywhere on $\left[a,b\right]$. Then we construct equivalence classes $\tilde{f}=\{g\in X:f\cong g\}$, and the vector space of these equivalence classes is $L^{2}[a,b]$, on which we define the norm $||\tilde{f}||_{1}=\sqrt{\int^{b}_{a}\left|f\left(x\right)\right|^{2}dx}$ (Lebesgue integral). Now some of these equivalences classes are rather special: they contain a continuous function in them, so this is the natural choice for a representative of the equivalence class. Let
$D\subseteq L^{2}[a,b]$ be the subspace containing these special equivalence classes. My basic question is, if we assign the equivalence classes in $D$ their continuous representatives, what are the natural representatives of the other equivalence classes?

We can make this more precise. Let $C[a,b]$ be the vector space of continuous functions $f:\left[a,b\right]\rightarrowℝ$, endowed with a norm $||f||_{2}=\sqrt{\int^{b}_{a}\left|f\left(x\right)\right|^{2}dx}$ (Riemann or Lebesgue integral). Then the norm-completion of this space is in fact $L^{2}[a,b]$. The upshot of all this is that $D$ is dense in $L^{2}[a,b]$, and we have a norm-respecting isomorphism $T:(D,||\cdot||_{1})\rightarrow(C[a,b] , ||\cdot||_{2})$ defined by $T(\tilde{f})\in \tilde{f}$ (assigning each element of $D$ its continuous representative). So now the question becomes, does there exist a continuous linear extension $S$ of $T$ defined on all of $L^{2}[a,b]$ such that $S|_{D}=T$ and $S(\tilde{f})\in \tilde{f}$ ? Well, $T$ is a bounded linear transformation (with operator norm 1) defined on a dense subspace, so it meets all the conditions of the BLT theorem other than the fact that its codomain is not a Banach space. Thus we have to expand $C[a,b]$ to a larger subspace of $X$, so that the codomain of $T$ becomes complete.

There are two potential ways to do this, depending on whether we define the norm $||\cdot||_{2}$ in terms of Riemann or Lebesgue integrals. If we use Riemann integrals, we would need a subspace of $X$ consisting of Riemann-integrable functions, so we would have to answer the following in order to establish completeness: if $f_{n}\rightarrow f$ with respect to the the $||\cdot||_{2}$ (where $f$ need not be continuous), is $f$ necessarily Riemann integrable? (My first instinct is no, because Riemann-integrability requires boundedness, and you can have a sequence of continuous functions with ever-increasing bounds, so that the limit is unbounded). If we use Lebesgue integrals, we would need to ensure that two distinct elements of the subspace cannot have zero distance, so we would have to answer the following: if $f_{n}\rightarrow f$ and $g_{n}\rightarrow g$ with respect to the $||\cdot||_{2}$ norm (where $f$ and $g$ need not be continuous) and $f(x)=g(x)$ almost everywhere on $[a,b]$, then are $f$ and $g$ necessarily the same function? (Again I fear the answer is no, because perhaps you can have a sequence of continuous functions that converges to a function with a removable discontinuity).

I know I've included a lot of convoluted detail, but my fundamental question is relatively simple: can we replace the equivalence classes in $L^{2}[a,b]$ with natural representative functions, using continuous representatives where possible? Or to put it another way: does there exist a subspace $Y$ of $X$ containing $C[a,b]$, on which we can define a norm which will make it isomorphic to $L^{2}[a,b]$?

EDIT: As Gerald has pointed out, a simpler way to phrase my question is that I want a lifting of $L^{2}[a,b]$ or more generally $L^{2}(ℝ^{3})$.

Even for the slightly more general case of equivalence classes containing piecewise continuous functions I don't see a natural way to proceed. How do we decide between $1_{(0, 1]}$ and $1_{[0, 1)}$ for example?
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Qiaochu YuanJan 1 '12 at 0:03

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Depending on one's ulterior goals, it might be useful to relate this to Sobolev space business. Let's look at the circle, instead of $[a,b]$, to dodge endpoint issues. Then $L^2(S^1)$ consists of constants + image of Sobolev space $H^2(S^1)$ under $d^2/dx^2$. By Sobolev imbedding/inequality, $H^2(S^1)$ is contained in $C^o(S^1)$, so functions in the "space of equivalence classes" $H^2(S^1)$ have a unique continuous representative. (Not all continuous functions are in $H^2$...) Then $L^2(S^1)$ is the image, plus constants. (If this direction is of interest, it is easy to elaborate...)
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paul garrettJan 1 '12 at 0:10

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Yemon, I mean something much more restrictive. I want $Y$ to be endowed with the specific norm $||f||=\sqrt{\int^{b}_{a}\left|f\left(x\right)\right|^{2}dx}$, but I'm open to this integral being either Riemann or Lebesgue.
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Keshav SrinivasanJan 1 '12 at 2:28

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Paul, I don't think I know enough to know whether or not Sobolev spaces is the direction I want to go in, but I can tell you my ulterior motive: quantum mechanics. $L^{2}(ℝ^{3})$ is a space of equivalence classes, but in QM you need to actually evaluate wavefunctions at points, so you need to choose representatives. In most common situations, the wavefunction is required to be continuous, so you choose a continuous representative out of an equivalence class that has one. But if you don't have continuity (which is possible!), the question becomes how can you choose a representative?
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Keshav SrinivasanJan 1 '12 at 16:27

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I would imagine that in practice it is not actually necessary to evaluate wave functions at every single point in order to do quantum mechanics. For instance, one can often proceed by interpreting all the equations of quantum mechanics in a distributional sense rather than a pointwise sense (note that the theory of distributions are very well adapted to linear PDE of the type encountered in QM). In many cases, the formal computations that appeared to require some regularity hypotheses can often be extended to the distributional setting by duality or a limiting argument.
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Terry TaoJan 2 '12 at 18:44

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The yes part is the Zorn lemma: consider the set of all subspaces $L\supset C[a,b]$ in the vector space of measurable square integrable functions such that no two functions in $L$ are equivalent partially ordered by inclusion. Since the union of any linearly ordered chain of such subspaces is such subspace again, we have a maximal such subspace $L$. It is easy to check that each square integrable function $f$ is equivalent to some function in $L$ (otherwise $\text{span\,}(L,f)$ is a bigger subspace).

The no part has been spelled out by Simon: no such subspace is any more reasonable or easier to put one's hands on than the Hamel basis of $\mathbb R$.

Fedja, what guarantee is there that $L$ is complete? If you have a sequence of continuous functions which converges to a square-integrable function $f$ (in the L2 norm), what guarantee is there that $f$ is actually in $L$, not just equivalent to some function in $L$?
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Keshav SrinivasanJan 3 '12 at 23:06

What do you mean? Equivalent functions are indistinguishable in $L^2$ and the limit in $L^2$ is defined up to a set of measure $0$. So, the function in $L$ that is the unique representative of the corresponding limit class is a limit of the sequence in the $L^2$-norm. Pointwise convergence has nothing to do with it.
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fedjaJan 4 '12 at 1:02

Small remark: Simon has withdrawn his original statement, although it seems like it should still be true
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Yemon ChoiJan 7 '12 at 2:44

A "lifting" is exactly a choice of one element of each equivalence class. When done on $L^\infty$, you want not only linear combinations of representatives to be representatives, but also products. There is a literature on this question. For example:

Gerald, I think Tulcea's text is out of print. Do you know whether lifting is possible for $L^{2}$ is possible if we only require that linear combination of representatives yields representatives?
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Keshav SrinivasanJan 3 '12 at 23:18

One way to partially answer your last question might be the following. To each $f\in L^2(a,b)$, first associate its Lebesgue primitive $F(x)=\int_a ^x f(t)dt$, then define $Tf$ as one of the four Dini derivatives of $F$, e.g.
$$ Tf(x)=\limsup _{h\to 0^+}h^{-1}(F(x+h)-F(x)).$$
Then $Tf=Tg$ everywhere if $f=g$ almost everywhere, $Tf=f$ almost everywhere, and $Tf$ is continuous if $f$ is equivalent to a continuous function. Thus the map $T$ associates to all members of a class of equivalence in $L^2$ the same function, which is the continuous representative of the class when it exists. An additional advantage is that the method is 'constructive'.

@ Piero D'Ancona: It seems to me that the construction you give does not make $T$ linear since $T(-f)=-Tf$ need not hold. Consider for example $f:t\mapsto 1+\sin(t^{-1})$ on $[0,1]$ with $f(0)=0$. Would taking the average of the upper and lower Dini derivatives be a cure?
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TaQJan 2 '12 at 18:06

Piero, assuming the linearity issue raised by TaQ can be resolved, this looks promising. But can Dini derivatives be defined for functions of more than one variable? I was just considering $L^{2}[a,b]$ for simplicity, but what I'm really interested in is $L^{2}(ℝ^{3})$. How would you define $T$ for that? Perhaps you could use the Hardy-Littlewood maximal operator defined here: en.wikipedia.org/wiki/Hardy-Littlewood_maximal_operator But you again face the issue: Wikipedia claims it's nonlinear.
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Keshav SrinivasanJan 3 '12 at 0:41

Actually, it seems to me that defining $T$ on $L^2[0,1]$ by $f\mapsto\frac12(D^+\int_0f+D_+\int_0f)$ does not make it linear either. Here $(\int_0f)(t)=\int_0^tf$, and $D^+$ and $D_+$ denote the upper and lower Dini derivatives from the right. But, as Terry Tao and Dmitri Pavlov have already pointed out, I also agree with the opinion that there is no need to get point values for the purpose of quantum mechanics.
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TaQJan 3 '12 at 18:41

I wonder if you could use the Lebesgue differentiation theorem. Let $F(B)$ be the Lebesgue integral of $F$ over the open ball $B$, and let $Tf(x)=\lim{|B|\to 0}\frac{F(B)}{|B|}$ where the limit is taken over open balls centered at $x$. Then $Tf=f$ almost everywhere, and thus $T$ chooses a unique representative out of each equivalence class, but does $T$ assign continuous representatives to the equivalence classes that have them, and is $T$ linear? If that doesn't work, is there anything we can do with the Hardy-Littlewood maximal operator? Does anyone know why it's not linear?
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Keshav SrinivasanJan 3 '12 at 23:55

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Then you do not get a vector space of functions since $+\infty$ and $-\infty$ cannot be added.
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TaQJan 5 '12 at 18:45

Well, there is a general sense in which your question can be answered in the affirmative. X = L^2 is a Banach space, and every Banach space X can be represented linearly and isometrically as a subspace of the continuous functions on a compact Hausdorff space K. The points of K are the continuous linear functionals on X. You deal with point functopns, not equivalence classes, but you have greatly extended the space of points.

I suppose this makes sense from a quantum mechanics standpoint; $X$ can consists of the vectors in the ket space, and $K$ can consist of normalized vectors (or "rays") in the bra space.
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Keshav SrinivasanJan 3 '12 at 1:27